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two positive point charges of 12uc and 8uc are 10cm apart. the work done in bringing them 4cm closer is?

tushar pal , 9 Years ago
Grade 12th pass
anser 6 Answers
Sourabh Singh

Last Activity: 9 Years ago

Hii

Find out the change in the potential energy of the system and take the difference. You will get the answer.


Best of Luck
Adarsh srivastav

Last Activity: 7 Years ago

q1=12×10^-6,q2=8×10^-6.
Wkt,potential when the are at R1(10 cm)
V1=kq1q2/R1=86.4×10^-1~=86×10^-1.
V2=kq1q2/R2=216×10^-1.
W.D=V2-V1=(216×10^-1)-(86×10^-1)=130×10^-1
                                                                =13J
Purba bandyopadhyay

Last Activity: 7 Years ago

So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta

Last Activity: 5 Years ago

 
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta

Last Activity: 5 Years ago

HiiFind out the change in the potential energy of the system and take the difference. You will get the answer.Best of Luck
 
Kushagra Madhukar

Last Activity: 5 Years ago

Dear student,
Please find the solution to your problem below.
 
Initial distance between the charges, d1 = 10 cm = 0.1 m
Final distance between the charges, d2 = 10 cm – 4 cm = 6 cm = 0.06 m
Hence, W = 9 x 109 x 12 x 10-6 x 8 x 10-6 (1/0.06 – 1/0.1)
               = 5.76 J
 
Hope it helps.
Thanks and regards,
Kushagra
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